# Ex.5.4 Q1 Arithmetic progressions Solutions - NCERT Maths Class 10

## Question

Which term of the A.P. \(121, 117, 113 ... \)is its first negative term?

[**Hint:** Find \(n\) for \({a_n} < 0\)]

## Text Solution

**What is Known?**

A.P. Series: \(121,\, 117, \,113 ...\)

**What is Unknown?**

First negative term of the AP

**Reasoning:**

\(n\rm{th}\) term of an AP is \(\,{a_n} = a + \left( {n - 1} \right)d.\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

**Steps:**

Given:

- First Term, \(a = 121\)
- Common difference, \(d = 117 - 121 = - 4\)

We know that \(n^\rm{th}\) term of AP, \({a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}{a_n} &= 121 + \left( {n - 1} \right) \times \left( { - 4} \right)\\ &= 121 - 4n + 4\\ &= 125 - 4n\end{align}\]

For the first negative term of this A.P,

\[\begin{align}{a_n} &< 0\\125 - 4n &< 0\\125 &< 4n\\n &> \frac{{125}}{4}\\n &> 31.25\end{align}\]

Therefore, \(32\rm{nd}\) term will be the first negative term of this A.P.